Problem: You have found the following ages (in years) of all 6 seals at your local zoo: $ 6,\enspace 15,\enspace 13,\enspace 2,\enspace 9,\enspace 2$ What is the average age of the seals at your zoo? What is the standard deviation? You may round your answers to the nearest tenth.
Explanation: Because we have data for all 6 seals at the zoo, we are able to calculate the population mean $({\mu})$ and population standard deviation $({\sigma})$ To find the population mean , add up the values of all $6$ ages and divide by $6$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{6}} x_i}{{6}} $ $ {\mu} = \dfrac{6 + 15 + 13 + 2 + 9 + 2}{{6}} = {7.8\text{ years old}} $ Find the squared deviations from the mean for each seal. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $6$ years $-1.8$ years $3.24$ years $^2$ $15$ years $7.2$ years $51.84$ years $^2$ $13$ years $5.2$ years $27.04$ years $^2$ $2$ years $-5.8$ years $33.64$ years $^2$ $9$ years $1.2$ years $1.44$ years $^2$ $2$ years $-5.8$ years $33.64$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{3.24} + {51.84} + {27.04} + {33.64} + {1.44} + {33.64}} {{6}} $ $ {\sigma^2} = \dfrac{{150.84}}{{6}} = {25.14\text{ years}^2} $ As you might guess from the notation, the population standard deviation $({\sigma})$ is found by taking the square root of the population variance $({\sigma^2})$ ${\sigma} = \sqrt{{\sigma^2}}$ $ {\sigma} = \sqrt{{25.14\text{ years}^2}} = {5\text{ years}} $ The average seal at the zoo is 7.8 years old. There is a standard deviation of 5 years.